LeetCode - 63. Unique Paths II

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博客围绕63. Unique Paths II展开，介绍了递归（会超时）、DP + 记忆化、DP二维数组、DP一维数组等解法，还提醒DP二维数组解法要从索引1开始。

63. Unique Paths II

Recursion (TLE)

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        return self.helper(obstacleGrid, len(obstacleGrid)-1, len(obstacleGrid[0])-1)
    def helper(self, grid, r, c):
        if grid[r][c]==1: return 0
        if r==0 and c==0: return 1
        if r==0: return self.helper(grid, r, c-1)
        if c==0: return self.helper(grid, r-1, c)
        return self.helper(grid, r, c-1)+self.helper(grid, r-1, c)

DP + memoization

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        mem = [[None for x in range(n)] for y in range(m)]
        return self.helper(obstacleGrid, m-1, n-1, mem)
    def helper(self, grid, r, c, mem):
        if grid[r][c]==1: return 0
        if r==0 and c==0: return 1
        if mem[r][c]==None:
            if r==0: temp = self.helper(grid, r, c-1, mem)
            elif c==0: temp = self.helper(grid, r-1, c, mem)
            else: temp = self.helper(grid, r-1, c, mem) + self.helper(grid, r, c-1, mem)
            mem[r][c] = temp
        return mem[r][c]

DP 2D array

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        mem = [[None for x in range(n)] for y in range(m)]
        for i in range(m):
            for j in range(n):
                if obstacleGrid[i][j]==1:
                    mem[i][j] = 0
                    continue
                if i==0 and j==0:
                    mem[i][j] = 1
                    continue
                if i==0: mem[i][j] = mem[i][j-1]
                elif j==0: mem[i][j] = mem[i-1][j]
                else: mem[i][j] = mem[i][j-1] + mem[i-1][j]
        return mem[m-1][n-1]

Attention: start from index=1

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        dp = [[0 for i in range(n+1)] for j in range(m+1)]
        dp[0][1] = 1
        for i in range(1,m+1):
            for j in range(1,n+1):
                if not obstacleGrid[i-1][j-1]:
                    dp[i][j] = dp[i][j-1]+dp[i-1][j]
        return dp[m][n]

DP 1D array

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        mem = [None for x in range(n)]
        for i in range(m):
            for j in range(n):
                if obstacleGrid[i][j]==1:
                    mem[j] = 0
                    continue
                if i==0 and j==0:
                    mem[j] = 1
                    continue
                if i==0: mem[j] = mem[j-1]
                elif j==0: continue
                else: mem[j] = mem[j-1] + mem[j]
        return mem[n-1]

class Solution:
    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
        m, n = len(obstacleGrid), len(obstacleGrid[0])
        mem = [0 for x in range(n)]
        mem[0] = 1
        for row in obstacleGrid:
            for j in range(n):
                if row[j]==1:
                    mem[j] = 0
                elif j>0:
                    mem[j] += mem[j-1]
        return mem[n-1]